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Salt of Weak Acid and Strong Base

100 mL of a 0...

Question

$100 m L$ of a $0.1 M C H_{3} C O O N a (a q)$ is titrated against a $0.1 M H C l (a q)$ . Calculate the $p H$ of the solution when $50 m L$ of $H C l$ was added.
Given: $p K_{a} (C H_{3} C O O H) = 4.74 a t 25^{o} C$

6.21

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5.85

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5.24

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4.74

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Solution

The correct option is D 4.74
mmol of $C H_{3} C O O N a = 100 \times 0.1 = 10$
mmol of $H C l = 50 \times 0.1 = 5$

$C H_{3} C O O N a (a q) + H C l (a q) \to C H_{3} C O O H (a q) + N a C l (l) I n i t i a l : 10500 F i n a l : 5055$

After titration equal milimoles of $C H_{3} C O O H$ and $C H_{3} C O O N a$ will form. The resultant solution will acts as an acidic buffer.
$Final concentration = \frac{Moles}{Total Volume}$
Total Volume $= 100 + 50 = 150 m L$
$[C H_{3} C O O H]_{f i n a l} = \frac{5}{150} M$
$[C H_{3} C O O N a]_{f i n a l} = \frac{5}{150} M$
The formula to calculate the $p H$ of an acidic buffer is :
$p H = p K_{a} + log (\frac{[C H_{3} C O O N a]}{[C H_{3} C O O H]}) p H = 4.74 + log ⎛ ⎝ \frac{\frac{5}{150}}{\frac{5}{150}} ⎞ ⎠ p H = 4.74$

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